Question

高中函数题，急

设函数f(x)=(sinwx+coswx)平方+2cos平方wx(w>0)的最小正周期为2π/3
1）求w的值
2）若函数y=g(x)的图像是由y=f（x）的图像向右平移π/2个单位长度所得到
求：y=g(x)的单调递增区间

Answer 1

f(x)=(sinwx+coswx)^2+2(coswx)^2
f(x)=1+2sinwxcoswx+2(coswx)^2
f(x)=1+sin2wx+cos2wx+1
f(x)=sin2wx+cos2wx+2
f(x)=√2(sin(2wx+π/4)+2
2π/3=2π/(2w)
w=3/2

g(x)=√2(sin(3(x-π/2)+π/4)+2
g(x)=√2(sin(3x-5π/4)+2

2kπ-π/2=<3x-5π/4<=2kπ+π/2
2/3kπ-π/4=<x<=2/3kπ+7π/12单调递增区间